Algebra - twice as much as other two combined

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$12$ tonnes of cement must be distributed among $3$ factories such that the first factory receives twice as much as other $2$ factories combined. How much cement does the first factory receive?

For this, I have come up with algebraic equation .

$x+y+z = 12$

$2(y+z)+y+z = 12$ (as per the statement above)

As the question did not say anything about other $2$ factories, we cannot assume that each of the other factories get equal amount of cement. Hence, I used independent values for the same.

and I know the answer for $x = 8$, so that $2(2+2) + 2 + 2 = 12$ through mental calculation.

But I don't know how to derive it mathematically. Can someone help me with this?

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I think you need to insert the step that you implied, to see how this works properly.

$x+y+z=12\quad$ (you have this)
$x=2(y+z)\quad$ (you implied this)

then there are various ways to go; perhaps double the first equation to produce a "$2(y+z)$":

$2x+2(y+z) = 24$
$3x = 24$
$ x= 8$

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Continuing from your attempt at a solution. Let $x$ be the amount of cement the first factory receives, $y$ and $z$ the amount that the second and third factory receive respectively.

The first factory receives twice as much as other two factories combined

This piece of information means that $x=2(y+z)$ hence we have the system of equations $$ \begin{align*} x+y+z&=12\tag{1}\\ x&=2(y+z)\tag{2} \end{align*} $$ Substituting (2) into (1) we get that $$ 3(y+z)=12\iff y+z=4.\tag{3} $$ Substituting (3) into one we get that $$ x+4=12\iff x=8. $$

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$x=$tonnes in first

$a=$tonnes in second + third

You obtain these two equations, since you only need to solve for the first factory:

$$$$ $$x=2a$$ $$x+a=12000$$

Solve for $x$ to determine the amount of tonnes of the first factory.

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The statement of the problem asks ONLY for the amount received by the first factory so let F (in tonnes) be that amount. Since the first factory receives twice as much as the other two combined, the other two receive a total of $F/2$. Since a total of $12$ tonnes is given to all three factories: $F + F/2 = 12$. Multiplying this equation by two gives $2F + F = 24$, so $3F = 24$ and thus $F = 8$.

Since the problem never asks how much each either of the other two factories receive there is no need to introduce variables for quantities not asked for, nor is it necessary to make any assumptions such quantities.

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Just expand out $2(y+y) + y + z = 12$.

$2(y+z) + y + z = (2y + 2z) +y + z = 3y + 3z = 3(y+z)$

So $3(y+z) = 12$

so $y + z = 4$.

So going back: $x + (y+z) = x + 4 = 12$

So $x = 8$.